71
Vivek @ Born this Way
·2010-09-11 03:52:00
equipotential plane ? Does it mean it returns to same level or height? :P
By the way, You in Edudigm?
49
Subhomoy Bakshi
·2010-09-11 07:03:16
yea me in edudigm...
equipotential plane here means the plane joining all the points having same gravitational potential!!
71
Vivek @ Born this Way
·2010-09-11 09:46:07
Of course, If they at same height then only they are in equipotential plane?
49
Subhomoy Bakshi
·2010-09-11 10:53:02
here there is no frame of reference like earth to define height...
but yeah your bhawna is correct!! ;)
@vivek: so everyone is nowadays copying rajiv bhaiya's status??? :P
71
Vivek @ Born this Way
·2010-09-12 06:03:27
Hhaahaha!
here there is no frame of reference like earth to define height...
Let define the launch Plane as the reference frame?
1
varun.tinkle
·2010-10-09 01:02:48
it snothing but a sum with dirty integration assuming the nagle to be theta instead of 90-theta as it makes no apparent difference
\left|v \right|=\left|ucos\theta i +(usin\theta -gt)j \right|
thereofre distance travelled is
\left|v \right|dt=\left|ucos\theta i +(usin\theta -gt)j \right|dt
more appropriately we hv to calculate the distanc esperately when v(y) is +ve and when it is -ve
so the answer comes up to ...u^{2}sin\theta (2cos\theta +sin\theta )/g
now taking into account that we hv assumed that we assumed the angle t be angle 90-@
so we now take that into account
so answer comes up to
u^{2}cos\theta (2sin\theta +cos\theta )/g
isnt it.......?